You've already learned the basic
trig graphs. But just as you could make the basic quadratic x2,
more complicated, such as –(x
+ 5)2 – 3, so also
trig graphs can be made more complicated. We can transform and translate
trig functions, just like you transformed and translated
other functions in algebra.

Let's start with the basic sine function,
f(t) = sin(t).
This function has an amplitude of 1
because the graph goes one unit up and one unit down from the midline
of the graph. This function has a period of 2π
because the sine wave repeats every 2π
units. The graph looks like this:

Now let's look at g(t)
= 3sin(t):

Do you see that the graph is three times
as tall? The amplitude has changed from 1 to
3.
This is always true: Whatever number A
is multiplied on the trig function gives you the amplitude; in this case,
that number was 3.
So 0.5cos(t)
would have an amplitude of 1/2,
and –2cos(t)
would have an amplitude of 2 and
would also be flipped upside down.

Do you see that the graph is squished in
from the sides? Do you see that the sine wave is cycling twice as fast,
so its period is only half as long? This is always true: Whatever value
B
is multiplied on the variable, you use this value to find the period ω
(omega) of the trig function, according to
this formula:

general
period formula:

For sines and cosines (and their reciprocals),
the "regular" period is 2π,
so the formula is:

period
formula for sines & cosines:

For tangents and cotangents, the "regular"
period is π,
so the formula is:

period
formula for tangents & cotangents:

In the sine wave graphed above, the value
of B was
2.
(Sometimes the value of B inside
the function will be negative, which is why there are absolute-value
bars on the denominator.) The formula for sines and cosines says that
cos(3t)
will have a period of (2π)/3
= (2/3)π; on the other hand, the formula
for tangents and cotangents says that tan(t/2)
will have a period of (2π)/(1/2)
= 4π.

(Note: Different books use different letters
to stand for the period formula. In your class, use whatever your book
or instructor uses.)

Now let's looks at j(t)
= sin(t – π/3):

Do you see that the graph (in blue) is
shifted over to the right by π/3
units from the regular graph (in gray)? This is always true: If the argument
of the function (the thing you're plugging in to the function) is of the
form (variable) – (number), then the graph is shifted to the right
by that (number) of units; if the argument is of the form (variable) +
(number), then the graph is shifted to the left by that (number)
of units. So cos(t
+ π/4) would be shifted to the left
by π/4
units, and tan(t
– 2π/3) would be shifted to the right
by (2/3)π
units. This right-or-left shifting is called "phase shift".

Now let's look at k(t)
= sin(t) + 3:

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Do you see how the graph was shifted up
by threeunits? This is always true: If a number D
is added outside the function, then the graph is shifted up by that number
of units; if D is
subtracted, then the graph is shifted down by that number of units. So
cos(t) – 2
is the regular cosine wave, but shifted downward two units; and tan(t)
+ 0.6 is the regular tangent curve,
but shifted upward by 6/10
of a unit.

Putting it all together, we have the general
sine function, F(t)
= Asin(Bt – C) + D, where
A is
the amplitude, B gives
you the period, D gives
you the vertical shift (up or down), and C is
used to find the phase shift. Why don't you always just use C?
Because sometimes more is going on inside the function. Remember that
the phase shift comes from what is added or subtracted directly to the
variable. So if you have something like sin(2t
– π), the phase shift is notπ units!
Instead, you first have to isolate what's happening to the variable by
factoring: sin(2(t
– π/2)). Now you can see that the phase
shift will be π/2
units, not π units.
So the phase shift, as a formula, is found by dividing C by
B.

For F(t)
= Af(Bt – C) + D,
where f(t)
is one of the basic trig functions, we have:

A:
amplitude is A

B:
period is (2π)/|B|

C:
phase shift is C/B

D:
vertical shift is D

Let's see what this looks like, in practice,
because there's a way to make these graphs a whole lot easier than what
they show in the book....